Preheat, adiabatic saturation and reheat on 100% outdoor air: 22-Mec-B2 Environmental Control in Buildings, December 2018, Q2

Question 2: Preheat, adiabatic saturation and reheat on 100% outdoor air (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Space sensible heat loss / latent load	200 000 Btu h$^{-1}$ / negligible
Space condition (state R)	$75\,{}^{\circ}\text{F}$ db, 50% RH
Outdoor condition (state O)	saturated at $20\,{}^{\circ}\text{F}$
Ventilation air, 100% outdoor	7000 scfm
Adiabatic saturator temperature	$60\,{}^{\circ}\text{F}$
Plant order	preheat coil → adiabatic saturator → reheat coil → space

Find. The plant diagram, the supply air temperature, the five state points with their dry- and wet-bulb temperatures, the preheat and reheat coil duties in Btu/hr, and the humidification water make-up in US gpm.

Figure 2.1 (part a) — The plant. Because the process demands 100% outdoor air there is no return-air path: all 7000 scfm is drawn in at state O, conditioned through states 2 and 3, delivered at state S and exhausted from the space. Make-up water enters the saturator sump at the saturator temperature.

Approach. The room load is entirely sensible, so the supply air must carry the room's own humidity ratio and be delivered hot enough to offset 200 000 Btu/hr. Working backwards, the saturator temperature fixes the $60\,{}^{\circ}\text{F}$ thermodynamic wet-bulb line: the preheat coil raises the outdoor air at constant humidity ratio until it reaches that line, the saturator moves it along the line up to the required humidity ratio, and the reheat coil then raises it at constant humidity ratio to the supply temperature.

Convert the ventilation rate to a dry-air mass flow. "Standard" cubic feet are referred to standard air at $0.075$ lb ft$^{-3}$, so $$\dot m_{da}=7000\ \text{ft}^3\text{min}^{-1}\times 60\ \frac{\text{min}}{\text{h}}\times 0.075\ \frac{\text{lb}}{\text{ft}^3}=31\,500\ \text{lb}_{da}\,\text{h}^{-1}$$ Every duty below is this one mass flow multiplied by an enthalpy or humidity-ratio difference.
Fix the room and outdoor states. With $p = 14.696$ psia, $W = 0.621945\,p_w/(p-p_w)$ and $h = 0.240\,t + W(1061 + 0.444\,t)$. At the room, $p_{ws}(75) = 0.4300$ psia and $\phi = 0.50$ give $W_R = 0.00924$ lb/lb (64.7 grains per pound) and $h_R = 28.11$ Btu/lb$_{da}$. Outdoor air saturated at $20\,{}^{\circ}\text{F}$ has $W_O = 0.002144$ lb/lb and $h_O = 7.09$ Btu/lb$_{da}$; being saturated, its wet bulb equals its dry bulb at $20\,{}^{\circ}\text{F}$.
Part (b) — the supply temperature. The room adds no moisture, so $W_S = W_R$ and the whole room load appears as a dry-bulb difference. With $c_p = 0.240 + 0.444W_R = 0.2441$ Btu lb$^{-1}$°F$^{-1}$, $$t_S=t_R+\frac{q_s}{\dot m_{da}c_p}=75+\frac{200\,000}{31\,500\times 0.2441}$$ $$\boxed{\;t_S = 101.0\,{}^{\circ}\text{F dry bulb},\ W_S = 0.00924\ \text{lb/lb}\ (22.0\%\ \text{RH})\;}$$ The familiar rule of thumb agrees: $t_S = 75 + 200\,000/(1.10\times 7000) = 101.0\,{}^{\circ}\text{F}$, which it must, because the 1.10 coefficient is built from the same standard density used in step 1.
Locate state 2, the preheat coil exit. A preheat coil adds no moisture, so state 2 lies on the vertical $W_O = 0.002144$ line; the saturator can only move air along a $60\,{}^{\circ}\text{F}$ wet-bulb line, so state 2 must already be on that line. Saturated air at the saturator temperature has $W^{*} = W_{sat}(60) = 0.011041$ lb/lb and $h^{*} = 26.41$ Btu/lb, and the adiabatic-saturation energy balance (the sump water enters at $t^{*}$, so $h_f = t^{*}-32 = 28$ Btu/lb) gives the enthalpy anywhere on that line as $$h = h^{*}-(W^{*}-W)\,h_f \quad\Rightarrow\quad h_2 = 26.41-(0.011041-0.002144)(28)=26.16\ \text{Btu/lb}$$ Inverting $h = 0.240t + W(1061+0.444t)$ at $W = W_O$, $$t_2=\frac{h_2-1061\,W_O}{0.240+0.444\,W_O}=\frac{26.16-2.275}{0.24095}=99.1\,{}^{\circ}\text{F}$$ so the preheat coil must lift the outdoor air all the way to $99.1\,{}^{\circ}\text{F}$ before the washer will do any good. Its relative humidity there is only 5.5% — hot, bone-dry air, which is precisely what an adiabatic humidifier needs.
Locate state 3, the saturator exit. The saturator must deliver the room's humidity ratio, $W_3 = W_R = 0.00924$ lb/lb. Moving along the same line, $h_3 = h_2 + (W_3-W_O)h_f = 26.16 + (0.007096)(28) = 26.36$ Btu/lb, and inverting the enthalpy relation at $W_3$ gives $$\boxed{\;t_3 = 67.8\,{}^{\circ}\text{F dry bulb},\ 63.7\%\ \text{RH},\ t_{wb}=60.0\,{}^{\circ}\text{F}\;}$$ The cabinet is therefore not saturating the air fully; its saturating efficiency is $$\eta_{sat}=\frac{W_3-W_O}{W^{*}-W_O}=\frac{0.007096}{0.008897}=79.7\%$$ which is a realistic single-bank spray performance and confirms that the plant as specified is physically achievable.

Part (c) — the five state points. Plotting states O, 2, 3, S and R on the IP chart gives the cycle in Figure 2.2, with dry- and wet-bulb temperatures as tabulated:

Point	Where	Dry bulb (°F)	Wet bulb (°F)	$W$ (lb/lb)	$h$ (Btu/lb)
O	outdoor air, saturated	20.0	20.0	0.002144	7.09
2	preheat coil exit	99.1	60.0	0.002144	26.16
3	saturator exit	67.8	60.0	0.00924	26.36
S	supply to space	101.0	70.8	0.00924	34.46
R	room / exhaust	75.0	62.6	0.00924	28.11

The signature of the cycle is that states 2 and 3 share a wet bulb of exactly $60\,{}^{\circ}\text{F}$ — that is what "adiabatic" means on this chart.

Figure 2.2 (part c) — The cycle on the IP chart. O→2 is horizontal (preheat, no moisture added); 2→3 runs down the $60\,{}^{\circ}\text{F}$ wet-bulb line (adiabatic saturation); 3→S is horizontal again (reheat); S→R is the sensible room process.

Part (d) — preheat coil duty. The coil raises the air from O to 2 at constant humidity ratio: $$Q_{pre}=\dot m_{da}(h_2-h_O)=31\,500\,(26.16-7.09)$$ $$\boxed{\;Q_{pre}= 600\,600\ \text{Btu}\,\text{h}^{-1}\;(176\ \text{kW})\;}$$ This is three times the room load, which is the honest cost of a 100%-outdoor-air process plant on a $20\,{}^{\circ}\text{F}$ day.
Part (e) — reheat coil duty. From state 3 to state S, again at constant humidity ratio: $$Q_{re}=\dot m_{da}(h_S-h_3)=31\,500\,(34.46-26.36)$$ $$\boxed{\;Q_{re}= 255\,100\ \text{Btu}\,\text{h}^{-1}\;(74.8\ \text{kW})\;}$$ Checking sensibly, $1.10\times 7000\times(101.0-67.8) = 255\,600$ Btu/hr, agreeing to 0.2%.
Part (f) — humidification water. The water evaporated is the humidity-ratio rise across the saturator carried on the dry-air flow: $$\dot m_w=\dot m_{da}(W_3-W_O)=31\,500\,(0.00924-0.002144)=223.5\ \text{lb}\,\text{h}^{-1}$$ Converting with the density of water at $60\,{}^{\circ}\text{F}$, 8.337 lb per US gallon, $$\boxed{\;\dot V_w = \frac{223.5}{8.337\times 60}=0.447\ \text{US gpm}\;(27\ \text{US gal per hour})\;}$$
Close the plant on an energy balance. Everything entering the air between O and S must come from the two coils plus the enthalpy of the make-up water: $$\dot m_{da}(h_S-h_O)-\dot m_{da}(W_3-W_O)h_f = 31\,500(34.46-7.09)-223.5(28)=855\,700\ \text{Btu}\,\text{h}^{-1}$$ and $Q_{pre}+Q_{re} = 600\,600+255\,100 = 855\,700$ Btu/hr. The two agree exactly, which confirms both coil duties and the water rate together.

Result	Value
Dry-air mass flow	31 500 lb$_{da}$ h$^{-1}$
(b) Air temperature entering the space	$101.0\,{}^{\circ}\text{F}$ db ($t_{wb}=70.8\,{}^{\circ}\text{F}$)
(c) Preheat exit, state 2	$99.1\,{}^{\circ}\text{F}$ db / $60.0\,{}^{\circ}\text{F}$ wb
(c) Saturator exit, state 3	$67.8\,{}^{\circ}\text{F}$ db / $60.0\,{}^{\circ}\text{F}$ wb (63.7% RH)
Saturator saturating efficiency	79.7%
(d) Preheat coil duty	600 600 Btu h$^{-1}$
(e) Reheat coil duty	255 100 Btu h$^{-1}$
(f) Humidification water	0.447 US gpm (223.5 lb h$^{-1}$)
Total plant heat input	855 700 Btu h$^{-1}$ (4.3× the room load)

Check: what "the temperature of the adiabatic saturator" means. It is read here as the saturator's adiabatic-saturation temperature — equal to the thermodynamic wet-bulb temperature of the air everywhere inside the device, and to the sump-water temperature — so states 2 and 3 both lie on the $60\,{}^{\circ}\text{F}$ wet-bulb line. That is the reading which makes the device adiabatic, and it is the only one under which the phrase describes the saturator rather than one of its air streams. The alternative reading, that the air leaves at $60\,{}^{\circ}\text{F}$ dry bulb, is rejected because at the required humidity ratio that state is only 84% saturated while sitting on a $57.0\,{}^{\circ}\text{F}$ wet-bulb line, so the cabinet would not be an adiabatic saturator at $60\,{}^{\circ}\text{F}$ at all. Stated as an assumption under cover-page instruction 1.

Topic. Winter plant psychrometry: sensible heating at constant humidity ratio, adiabatic (evaporative) humidification along a line of constant thermodynamic wet bulb, and the preheat/reheat split that a fixed room state forces.

Key relations. Humidity ratio $W = 0.621945\,p_w/(p-p_w)$; enthalpy $h = 0.240t + W(1061+0.444t)$ Btu/lb$_{da}$; adiabatic-saturation line $h = h^{*}-(W^{*}-W)h_f(t^{*})$; sensible load $q_s = \dot m_{da}c_p\Delta t \approx 1.10\,\text{cfm}\,\Delta t$; latent load $q_l \approx 4840\,\text{cfm}\,\Delta W$; saturating efficiency $\eta = (W_{out}-W_{in})/(W^{*}-W_{in})$; make-up water $\dot m_w = \dot m_{da}\Delta W$.

Why this works. An adiabatic humidifier exchanges no heat with its surroundings, so the only enthalpy crossing its boundary is the small amount carried by the liquid make-up water. That makes the process very nearly one of constant enthalpy, and exactly one of constant thermodynamic wet bulb — which is why the wet-bulb line, not the enthalpy line, is the correct construction. The consequence drives the whole design: the air must be preheated until its wet bulb reaches the saturator temperature, because the washer cannot raise wet bulb at all. Hence the surprisingly high $99.1\,{}^{\circ}\text{F}$ at state 2, and hence the fact that the saturator then cools the air by 31 °F while wetting it, so that a reheat coil is needed afterwards to put the supply temperature back up.

Common pitfall. The most common error is preheating only to the supply temperature and then expecting the washer to add the moisture — it cannot, because at $101\,{}^{\circ}\text{F}$ and $W_O$ the wet bulb is $60.6\,{}^{\circ}\text{F}$, so the air would overshoot the required humidity or need a warmer sump. The second is treating the saturator as a constant-enthalpy device and forgetting the $(W^{*}-W)h_f$ term; the error is only about 0.25 Btu/lb here but it moves state 2 by a degree and breaks the closing energy balance in step 10. The third is quoting the water make-up as a mass rate when the question asks for gpm, or converting with the density of water at $32\,{}^{\circ}\text{F}$ rather than at the sump temperature.

Source. ASHRAE Handbook — Fundamentals, Ch. 1, eqs. 20–33 and the adiabatic-saturation development; McQuiston, Parker & Spitler, HVAC Analysis and Design, 6th ed., §3.4–3.6 (moist-air processes) and §3.9 (humidification); Jones, Air Conditioning Engineering, 5th ed., Ch. 3.

Question 2 of 8: Preheat, adiabatic saturation and reheat on 100% outdoor air

Question 2: Preheat, adiabatic saturation and reheat on 100% outdoor air (10 marks)